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5-6 The Quadratic Formula
and the Discriminant
p.292-297
We have a number of different way
of finding the roots if a quadratic
equations
#1. Making a table
#2. Factoring
#3. Completing the Square
Now a new way that comes from completing
the square.
The Quadratic Formula
The Quadratic Formula
Solve for x by completing the square.
ax 2 + bx + c = 0
ax 2 + bx +  = −c + 
b
−c
2
x + x +=
+
a
a
2

b
−c  b 
 b 
x + x+  =
+ 
a
a  2a 
 2a 
2

2
The Quadratic Formula
Solve for x by completing the square.
ax 2 + bx + c = 0
ax 2 + bx +  = −c + 
b
−c
x2 + x +  =
+
a
a
2

b
−c  b 
 b 
x2 + x +   =
+ 
a
a  2a 
 2a 

b
b2
b 2 − 4ac
x2 + x + 2 =
a
4a
4a 2
2

2

b  b 2 − 4ac

x+  =
2a 
4a 2

The Quadratic Formula
Solve for x by completing the square.
b
b2
b 2 − 4ac
x + x+ 2 =
a
4a
4a 2
2

2

b  b 2 − 4ac

x+  =
2a 
4a 2


b
b 2 − 4ac
x+
=±
2a
4a 2
−b
b 2 − 4ac
x=
±
2a
2a
− b ± b 2 − 4ac
x=
2a
The Quadratic Formula
Solve for x by completing the square.

− b ± b − 4ac
x=
2a
2
How does it work
Equation:
3x 2 + 5 x + 1 = 0
a=3
b=5
c =1
− b ± b 2 − 4ac
x=
2a
How does it work
Equation:
3x + 5 x + 1 = 0
a=3
b=5
c =1
2

− b ± b 2 − 4ac
x=
2a

x=

− ( 5) ±

( 5) 2 − 4( 3)(1)
2( 3)

− 5 ± 25 − 12
x=
6
x=

− 5 ± 13 − 5
13
=
±
6
6
6
The Discriminant
The number in the square root of the
quadratic formula.

b − 4ac
2

Given x − 5 x + 6 = 0
2

( − 5)

− 4(1)( 6 )
25 − 24 = 1
2
The Discriminant

b − 4ac
2

The Discriminant can be negative, positive or zero
If the Discriminant is positive then there are:
2 real answers.
If the square root is not a prefect square
( for example 25 ),
then there will be 2 irrational roots
( for example 2 ± 5 ).
The Discriminant

b 2 − 4ac

The Discriminant can be negative, positive or zero
If the Discriminant is positive,
there are 2 real answers.
If the Discriminant is zero,
there is 1 real answer.
If the Discriminant is negative,
there are 2 complex answers.
complex answer have i.
Let’s put all of that
b
information in a chart.
Value of Discriminant

Type and
Number of Roots

D > 0,
D is a perfect square

2 real,
rational roots
(ex: x= 2 and x= -4)

D > 0,
D NOT a perfect square

2 real,
Irrational roots
(x = √13 x= -√13)

D=0

1 real, rational root
(double root)
(ex: x = 5)

D<0

2 complex roots
(complex conjugates)
(x = 2 ± 3i )

2

− 4ac

Sample Graph
of Related Function
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0

0, One rational root
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0

0, One rational root
x + 3x + 5 = 0
2

-11, Two complex roots
x + 8x − 4 = 0
2

80, Two irrational roots
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0

0, One rational root
x + 3x + 5 = 0
2
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0

0, One rational root
x + 3x + 5 = 0
2

-11, Two complex roots
Describe the roots
Tell me the Discriminant and the type of roots

x2 + 6x + 9 = 0

0, One rational root
x + 3x + 5 = 0
2

-11, Two complex roots
x + 8x − 4 = 0
2
Solve using the Quadratic formula

x − 8 x = 33
2
Solve using the Quadratic formula
x − 8 x = 33
2

x − 8 x − 33 = 0
2

x=

− ( − 8) ±

( − 8) − 4(1)( − 33)
2(1)
2
Solve using the Quadratic formula
x 2 − 8 x = 33
x 2 − 8 x − 33 = 0
x=

− ( − 8) ±

( − 8) 2 − 4(1)( − 33)
2(1)

8 ± 196 8 ± 14
x=
=
2
2
8 + 14 22
x=
=
= 11
2
2
8 − 14 − 6
x=
=
= −3
2
2
Solve using the Quadratic formula

x − 34 x + 289 = 0
2
Solve using the Quadratic formula
x − 34 x + 289 = 0
2

x=

− ( − 34 ) ±

( − 34)
2(1)

2

− 4(1)( 289 )
Solve using the Quadratic formula
x − 34 x + 289 = 0
2

x=

− ( − 34) ±

( − 34) 2 − 4(1)( 289)
2(1)

34 ± 1156 − 1156
x=
2
34 ± 0 34
x=
=
= 17
2
2
Solve using the Quadratic formula

x − 6x + 2 = 0
2
Solve using the Quadratic formula
x − 6x + 2 = 0
2

x=

− ( − 6) ±

( − 6)
2(1)

2

− 4(1)( 2 )

6 ± 36 − 8 6 ± 28
x=
=
2
2
6 2 7
x= ±
= 3± 7
2
2
Solve using the Quadratic formula

x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=

− ( − 6) ±

( − 6) 2 − 4(1)(13)
2(1)
Solve using the Quadratic formula

x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=

− ( − 6) ±

( − 6) 2 − 4(1)(13)
2(1)

6 ± 36 − 52 6 ± − 16
x=
=
2
2
Solve using the Quadratic formula
x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=

− ( − 6) ±

( − 6) 2 − 4(1)(13)
2(1)

6 ± 36 − 52 6 ± − 16
x=
=
2
2
6 ± 4i 6 4
x=
= ± i
2
2 2
x = 3 ± 2i

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Quadratic eq and discriminant

  • 1. 5-6 The Quadratic Formula and the Discriminant p.292-297
  • 2. We have a number of different way of finding the roots if a quadratic equations #1. Making a table #2. Factoring #3. Completing the Square Now a new way that comes from completing the square. The Quadratic Formula
  • 3. The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c 2 x + x += + a a 2 b −c  b   b  x + x+  = +  a a  2a   2a  2 2
  • 4. The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c x2 + x +  = + a a 2 b −c  b   b  x2 + x +   = +  a a  2a   2a  b b2 b 2 − 4ac x2 + x + 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2 
  • 5. The Quadratic Formula Solve for x by completing the square. b b2 b 2 − 4ac x + x+ 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2  b b 2 − 4ac x+ =± 2a 4a 2 −b b 2 − 4ac x= ± 2a 2a − b ± b 2 − 4ac x= 2a
  • 6. The Quadratic Formula Solve for x by completing the square. − b ± b − 4ac x= 2a 2
  • 7. How does it work Equation: 3x 2 + 5 x + 1 = 0 a=3 b=5 c =1 − b ± b 2 − 4ac x= 2a
  • 8. How does it work Equation: 3x + 5 x + 1 = 0 a=3 b=5 c =1 2 − b ± b 2 − 4ac x= 2a x= − ( 5) ± ( 5) 2 − 4( 3)(1) 2( 3) − 5 ± 25 − 12 x= 6 x= − 5 ± 13 − 5 13 = ± 6 6 6
  • 9. The Discriminant The number in the square root of the quadratic formula. b − 4ac 2 Given x − 5 x + 6 = 0 2 ( − 5) − 4(1)( 6 ) 25 − 24 = 1 2
  • 10. The Discriminant b − 4ac 2 The Discriminant can be negative, positive or zero If the Discriminant is positive then there are: 2 real answers. If the square root is not a prefect square ( for example 25 ), then there will be 2 irrational roots ( for example 2 ± 5 ).
  • 11. The Discriminant b 2 − 4ac The Discriminant can be negative, positive or zero If the Discriminant is positive, there are 2 real answers. If the Discriminant is zero, there is 1 real answer. If the Discriminant is negative, there are 2 complex answers. complex answer have i.
  • 12. Let’s put all of that b information in a chart. Value of Discriminant Type and Number of Roots D > 0, D is a perfect square 2 real, rational roots (ex: x= 2 and x= -4) D > 0, D NOT a perfect square 2 real, Irrational roots (x = √13 x= -√13) D=0 1 real, rational root (double root) (ex: x = 5) D<0 2 complex roots (complex conjugates) (x = 2 ± 3i ) 2 − 4ac Sample Graph of Related Function
  • 13. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0
  • 14. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root
  • 15. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2 80, Two irrational roots
  • 16. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2
  • 17. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots
  • 18. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2
  • 19. Solve using the Quadratic formula x − 8 x = 33 2
  • 20. Solve using the Quadratic formula x − 8 x = 33 2 x − 8 x − 33 = 0 2 x= − ( − 8) ± ( − 8) − 4(1)( − 33) 2(1) 2
  • 21. Solve using the Quadratic formula x 2 − 8 x = 33 x 2 − 8 x − 33 = 0 x= − ( − 8) ± ( − 8) 2 − 4(1)( − 33) 2(1) 8 ± 196 8 ± 14 x= = 2 2 8 + 14 22 x= = = 11 2 2 8 − 14 − 6 x= = = −3 2 2
  • 22. Solve using the Quadratic formula x − 34 x + 289 = 0 2
  • 23. Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34 ) ± ( − 34) 2(1) 2 − 4(1)( 289 )
  • 24. Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34) ± ( − 34) 2 − 4(1)( 289) 2(1) 34 ± 1156 − 1156 x= 2 34 ± 0 34 x= = = 17 2 2
  • 25. Solve using the Quadratic formula x − 6x + 2 = 0 2
  • 26. Solve using the Quadratic formula x − 6x + 2 = 0 2 x= − ( − 6) ± ( − 6) 2(1) 2 − 4(1)( 2 ) 6 ± 36 − 8 6 ± 28 x= = 2 2 6 2 7 x= ± = 3± 7 2 2
  • 27. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1)
  • 28. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2
  • 29. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2 6 ± 4i 6 4 x= = ± i 2 2 2 x = 3 ± 2i